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In an arithmetic sequence, the third term is 10 and the fifth term is 16.
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I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.
In an arithmetic sequence, the third term is $10$ and the fifth term is $16$.
Find the common difference.
Find the first term.
Find the sum of the first $20$ terms in the sequence.
So, the arithmetic formula is $a_n = a_1 + (n – 1)d$ right?
The common difference is the difference between the terms I think. So $16 - 10 = 6$, but there is a term between that so divided by $2$ it is $3$.
How do I find the first term and the sum of the first 20 terms?
sequences-and-series algebra-precalculus
edited Jul 31 at 15:57
N. F. Taussig
38k93053
asked Jul 31 at 15:04
Ella
587
add a comment |Â
up vote
3
down vote
favorite
I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.
In an arithmetic sequence, the third term is $10$ and the fifth term is $16$.
Find the common difference.
Find the first term.
Find the sum of the first $20$ terms in the sequence.
So, the arithmetic formula is $a_n = a_1 + (n – 1)d$ right?
The common difference is the difference between the terms I think. So $16 - 10 = 6$, but there is a term between that so divided by $2$ it is $3$.
How do I find the first term and the sum of the first 20 terms?
sequences-and-series algebra-precalculus
edited Jul 31 at 15:57
N. F. Taussig
38k93053
asked Jul 31 at 15:04
Ella
587
1
You've done well so far. If the common difference is $3$ and the third term is $10$, what is the second term? Once you answer that, you'll find the rest is easy.
– saulspatz
Jul 31 at 15:11
2
@saulspatz Oh! So, the second term is 7 and the first term would be 4.
– Ella
Jul 31 at 15:15
There you go, that's it!
– saulspatz
Jul 31 at 15:18
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.
In an arithmetic sequence, the third term is $10$ and the fifth term is $16$.
Find the common difference.
Find the first term.
Find the sum of the first $20$ terms in the sequence.
So, the arithmetic formula is $a_n = a_1 + (n – 1)d$ right?
The common difference is the difference between the terms I think. So $16 - 10 = 6$, but there is a term between that so divided by $2$ it is $3$.
How do I find the first term and the sum of the first 20 terms?
sequences-and-series algebra-precalculus
edited Jul 31 at 15:57
N. F. Taussig
38k93053
asked Jul 31 at 15:04
Ella
587
I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.
In an arithmetic sequence, the third term is $10$ and the fifth term is $16$.
Find the common difference.
Find the first term.
Find the sum of the first $20$ terms in the sequence.
So, the arithmetic formula is $a_n = a_1 + (n – 1)d$ right?
The common difference is the difference between the terms I think. So $16 - 10 = 6$, but there is a term between that so divided by $2$ it is $3$.
How do I find the first term and the sum of the first 20 terms?
sequences-and-series algebra-precalculus
edited Jul 31 at 15:57
N. F. Taussig
38k93053
asked Jul 31 at 15:04
Ella
587
edited Jul 31 at 15:57
N. F. Taussig
38k93053
edited Jul 31 at 15:57
N. F. Taussig
38k93053
edited Jul 31 at 15:57
N. F. Taussig
38k93053
38k93053
asked Jul 31 at 15:04
Ella
587
asked Jul 31 at 15:04
Ella
587
asked Jul 31 at 15:04
Ella
587
587
1
You've done well so far. If the common difference is $3$ and the third term is $10$, what is the second term? Once you answer that, you'll find the rest is easy.
– saulspatz
Jul 31 at 15:11
2
@saulspatz Oh! So, the second term is 7 and the first term would be 4.
– Ella
Jul 31 at 15:15
There you go, that's it!
– saulspatz
Jul 31 at 15:18
add a comment |Â
1
You've done well so far. If the common difference is $3$ and the third term is $10$, what is the second term? Once you answer that, you'll find the rest is easy.
– saulspatz
Jul 31 at 15:11
2
@saulspatz Oh! So, the second term is 7 and the first term would be 4.
– Ella
Jul 31 at 15:15
There you go, that's it!
– saulspatz
Jul 31 at 15:18
1
1
You've done well so far. If the common difference is $3$ and the third term is $10$, what is the second term? Once you answer that, you'll find the rest is easy.
– saulspatz
Jul 31 at 15:11
You've done well so far. If the common difference is $3$ and the third term is $10$, what is the second term? Once you answer that, you'll find the rest is easy.
– saulspatz
Jul 31 at 15:11
2
2
@saulspatz Oh! So, the second term is 7 and the first term would be 4.
– Ella
Jul 31 at 15:15
@saulspatz Oh! So, the second term is 7 and the first term would be 4.
– Ella
Jul 31 at 15:15
There you go, that's it!
– saulspatz
Jul 31 at 15:18
There you go, that's it!
– saulspatz
Jul 31 at 15:18
add a comment |Â
2 Answers
2
active
oldest
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up vote
2
down vote
The pedestrian approach is to use your formula $a_n=a_1+(n-1)d$ and plug in the two values you know
$$a_3=10=a_1+(3-1)d\a_5=16=a_1+(5-1)d$$
This is two simultaneous equations that you can solve for $a_1,d$. You have already found $d=3$ essentially by subtracting the two. Now just plug that into one of them and evaluate $a_1$.
Otherwise, since you know $d=3$ you can just count two steps down from $a_3=10$ to get $a_1$
To get the sum of the first $20$ terms you need to know (and should have proved) that $$sum_i=1^ni=frac 12n(n+1)$$ If you write out the sum of the first $20$ terms you will need $d$ times this with $n=19$ plus to account for the $a$s
answered Jul 31 at 15:18
Ross Millikan
275k21184351
You meant to write $$sum_i = 1^n i = frac12n(n + 1)$$ Also, check your last sentence, which seems to suggest that multiplying this sum by $d$ will yield the sum of the first $20$ terms of the arithmetic sequence.
– N. F. Taussig
Aug 1 at 17:19
@N.F.Taussig: Thanks, fixed, and a remark about the $a$s added
– Ross Millikan
Aug 1 at 19:53
I completed the whole problem and just wanted some clarification, if that was okay. For c, I got 650. Does this sound correct?
– Ella
Aug 4 at 2:44
Yes, that is correct.
– Ross Millikan
Aug 4 at 4:06
add a comment |Â
up vote
1
down vote
You have correctly found the common difference.
The $k$th term of an arithmetic sequence with initial term $a_1$ and common difference $d$ is
$$a_k = a_1 + (k - 1)d$$
Since we are given that $a_3 = 10$ and $a_5 = 16$, we obtain
beginalign*
a_3 & = a_1 + (3 - 1)d = 10\
a_5 & = a_1 + (5 - 1)d = 16
endalign*
which is a system of two linear equations in two variables. Subtracting the first equation from the second yields $2d = 6 implies d = 3$, as you found. Substituting $d = 3$ in the equation for $a_3$ yields $a_1 = 4$, as you found in the comments.
The $n$th partial sum (the sum of the first $n$ terms) of an arithmetic series
beginalign*
S_n & = sum_k = 1^n a_k\
& = sum_k = 1^n [a_1 + (k - 1)d]\
& = sum_k = 1^n a_1 + sum_k = 1^n (k - 1)d\
& = a_1sum_k = 1^n 1 + dsum_k = 1^n (k - 1)\
& = na_1 + fracd(n - 1)n2
endalign*
from which you can find $S_20$ by substituting $20$ for $n$, $3$ for $d$, and $4$ for $a_1$.
Alternatively,
beginalign*
S_n & = sum_k = 1^n a_k\
& = a_1 + a_2 + a_3 + cdots + a_n - 2 + a_n - 1 + a_n\
& = a_1 + [a_1 + d] + [a_1 + 2d] + cdots + [a_1 + (n - 3)d] + [a_1 + (n - 2)d] + [a_1 + (n - 1)d]
endalign*
Since we obtain the same sum if we write the terms in reverse order, we obtain
beginalign*
S_n & = a_n + a_n - 1 + a_n - 2 + cdots + a_3 + a_2 + a_1\
& = [a_1 + (n - 1)d] + [a_1 + (n - 2)d] + [a_1 + (n - 3)d] + cdots + [a_1 + 2d] + [a_1 + d] + a_1
endalign*
Adding the two expressions for $S_n$ yields
beginalignat*10
S_n & = & a_1 & + & [a_1 + d] & + & cdots & + & [a_1 + (n - 1)d] & + & [a_1 + (n - 1)d]\
S_n & = & [a_1 + (n - 1)d] & + & [a_1 + (n - 2)d] & + & cdots & + & [a_1 + d] & + & a_1\ hline
2S_n & = & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d] & + & cdots & + & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d]
endalignat*
Since there are $n$ columns in that sum, we obtain
beginalign*
2S_n & = n[2a_1 + (n - 1)d]\
2S_n & = n[a_1 + a_1 + (n - 1)d]\
2S_n & = n(a_1 + a_n)\
S_n & = fracn(a_1 + a_n)2
endalign*
from which you can find $S_20$ by substituting $4$ for $a_1$ and using the formula $a_n = a_1 + (n - 1)d$ with $n = 20$ and $d = 3$ to find $a_20$.
answered Aug 1 at 9:43
N. F. Taussig
38k93053
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The pedestrian approach is to use your formula $a_n=a_1+(n-1)d$ and plug in the two values you know
$$a_3=10=a_1+(3-1)d\a_5=16=a_1+(5-1)d$$
This is two simultaneous equations that you can solve for $a_1,d$. You have already found $d=3$ essentially by subtracting the two. Now just plug that into one of them and evaluate $a_1$.
Otherwise, since you know $d=3$ you can just count two steps down from $a_3=10$ to get $a_1$
To get the sum of the first $20$ terms you need to know (and should have proved) that $$sum_i=1^ni=frac 12n(n+1)$$ If you write out the sum of the first $20$ terms you will need $d$ times this with $n=19$ plus to account for the $a$s
answered Jul 31 at 15:18
Ross Millikan
275k21184351
You meant to write $$sum_i = 1^n i = frac12n(n + 1)$$ Also, check your last sentence, which seems to suggest that multiplying this sum by $d$ will yield the sum of the first $20$ terms of the arithmetic sequence.
– N. F. Taussig
Aug 1 at 17:19
@N.F.Taussig: Thanks, fixed, and a remark about the $a$s added
– Ross Millikan
Aug 1 at 19:53
I completed the whole problem and just wanted some clarification, if that was okay. For c, I got 650. Does this sound correct?
– Ella
Aug 4 at 2:44
Yes, that is correct.
– Ross Millikan
Aug 4 at 4:06
add a comment |Â
up vote
2
down vote
The pedestrian approach is to use your formula $a_n=a_1+(n-1)d$ and plug in the two values you know
$$a_3=10=a_1+(3-1)d\a_5=16=a_1+(5-1)d$$
This is two simultaneous equations that you can solve for $a_1,d$. You have already found $d=3$ essentially by subtracting the two. Now just plug that into one of them and evaluate $a_1$.
Otherwise, since you know $d=3$ you can just count two steps down from $a_3=10$ to get $a_1$
To get the sum of the first $20$ terms you need to know (and should have proved) that $$sum_i=1^ni=frac 12n(n+1)$$ If you write out the sum of the first $20$ terms you will need $d$ times this with $n=19$ plus to account for the $a$s
answered Jul 31 at 15:18
Ross Millikan
275k21184351
You meant to write $$sum_i = 1^n i = frac12n(n + 1)$$ Also, check your last sentence, which seems to suggest that multiplying this sum by $d$ will yield the sum of the first $20$ terms of the arithmetic sequence.
– N. F. Taussig
Aug 1 at 17:19
@N.F.Taussig: Thanks, fixed, and a remark about the $a$s added
– Ross Millikan
Aug 1 at 19:53
I completed the whole problem and just wanted some clarification, if that was okay. For c, I got 650. Does this sound correct?
– Ella
Aug 4 at 2:44
Yes, that is correct.
– Ross Millikan
Aug 4 at 4:06
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The pedestrian approach is to use your formula $a_n=a_1+(n-1)d$ and plug in the two values you know
$$a_3=10=a_1+(3-1)d\a_5=16=a_1+(5-1)d$$
This is two simultaneous equations that you can solve for $a_1,d$. You have already found $d=3$ essentially by subtracting the two. Now just plug that into one of them and evaluate $a_1$.
Otherwise, since you know $d=3$ you can just count two steps down from $a_3=10$ to get $a_1$
To get the sum of the first $20$ terms you need to know (and should have proved) that $$sum_i=1^ni=frac 12n(n+1)$$ If you write out the sum of the first $20$ terms you will need $d$ times this with $n=19$ plus to account for the $a$s
answered Jul 31 at 15:18
Ross Millikan
275k21184351
The pedestrian approach is to use your formula $a_n=a_1+(n-1)d$ and plug in the two values you know
$$a_3=10=a_1+(3-1)d\a_5=16=a_1+(5-1)d$$
This is two simultaneous equations that you can solve for $a_1,d$. You have already found $d=3$ essentially by subtracting the two. Now just plug that into one of them and evaluate $a_1$.
Otherwise, since you know $d=3$ you can just count two steps down from $a_3=10$ to get $a_1$
To get the sum of the first $20$ terms you need to know (and should have proved) that $$sum_i=1^ni=frac 12n(n+1)$$ If you write out the sum of the first $20$ terms you will need $d$ times this with $n=19$ plus to account for the $a$s
answered Jul 31 at 15:18
Ross Millikan
275k21184351
edited Aug 1 at 19:52
answered Jul 31 at 15:18
Ross Millikan
275k21184351
answered Jul 31 at 15:18
Ross Millikan
275k21184351
answered Jul 31 at 15:18
Ross Millikan
275k21184351
275k21184351
You meant to write $$sum_i = 1^n i = frac12n(n + 1)$$ Also, check your last sentence, which seems to suggest that multiplying this sum by $d$ will yield the sum of the first $20$ terms of the arithmetic sequence.
– N. F. Taussig
Aug 1 at 17:19
@N.F.Taussig: Thanks, fixed, and a remark about the $a$s added
– Ross Millikan
Aug 1 at 19:53
I completed the whole problem and just wanted some clarification, if that was okay. For c, I got 650. Does this sound correct?
– Ella
Aug 4 at 2:44
Yes, that is correct.
– Ross Millikan
Aug 4 at 4:06
add a comment |Â
You meant to write $$sum_i = 1^n i = frac12n(n + 1)$$ Also, check your last sentence, which seems to suggest that multiplying this sum by $d$ will yield the sum of the first $20$ terms of the arithmetic sequence.
– N. F. Taussig
Aug 1 at 17:19
@N.F.Taussig: Thanks, fixed, and a remark about the $a$s added
– Ross Millikan
Aug 1 at 19:53
I completed the whole problem and just wanted some clarification, if that was okay. For c, I got 650. Does this sound correct?
– Ella
Aug 4 at 2:44
Yes, that is correct.
– Ross Millikan
Aug 4 at 4:06
You meant to write $$sum_i = 1^n i = frac12n(n + 1)$$ Also, check your last sentence, which seems to suggest that multiplying this sum by $d$ will yield the sum of the first $20$ terms of the arithmetic sequence.
– N. F. Taussig
Aug 1 at 17:19
You meant to write $$sum_i = 1^n i = frac12n(n + 1)$$ Also, check your last sentence, which seems to suggest that multiplying this sum by $d$ will yield the sum of the first $20$ terms of the arithmetic sequence.
– N. F. Taussig
Aug 1 at 17:19
@N.F.Taussig: Thanks, fixed, and a remark about the $a$s added
– Ross Millikan
Aug 1 at 19:53
@N.F.Taussig: Thanks, fixed, and a remark about the $a$s added
– Ross Millikan
Aug 1 at 19:53
I completed the whole problem and just wanted some clarification, if that was okay. For c, I got 650. Does this sound correct?
– Ella
Aug 4 at 2:44
I completed the whole problem and just wanted some clarification, if that was okay. For c, I got 650. Does this sound correct?
– Ella
Aug 4 at 2:44
Yes, that is correct.
– Ross Millikan
Aug 4 at 4:06
Yes, that is correct.
– Ross Millikan
Aug 4 at 4:06
add a comment |Â
up vote
1
down vote
You have correctly found the common difference.
The $k$th term of an arithmetic sequence with initial term $a_1$ and common difference $d$ is
$$a_k = a_1 + (k - 1)d$$
Since we are given that $a_3 = 10$ and $a_5 = 16$, we obtain
beginalign*
a_3 & = a_1 + (3 - 1)d = 10\
a_5 & = a_1 + (5 - 1)d = 16
endalign*
which is a system of two linear equations in two variables. Subtracting the first equation from the second yields $2d = 6 implies d = 3$, as you found. Substituting $d = 3$ in the equation for $a_3$ yields $a_1 = 4$, as you found in the comments.
The $n$th partial sum (the sum of the first $n$ terms) of an arithmetic series
beginalign*
S_n & = sum_k = 1^n a_k\
& = sum_k = 1^n [a_1 + (k - 1)d]\
& = sum_k = 1^n a_1 + sum_k = 1^n (k - 1)d\
& = a_1sum_k = 1^n 1 + dsum_k = 1^n (k - 1)\
& = na_1 + fracd(n - 1)n2
endalign*
from which you can find $S_20$ by substituting $20$ for $n$, $3$ for $d$, and $4$ for $a_1$.
Alternatively,
beginalign*
S_n & = sum_k = 1^n a_k\
& = a_1 + a_2 + a_3 + cdots + a_n - 2 + a_n - 1 + a_n\
& = a_1 + [a_1 + d] + [a_1 + 2d] + cdots + [a_1 + (n - 3)d] + [a_1 + (n - 2)d] + [a_1 + (n - 1)d]
endalign*
Since we obtain the same sum if we write the terms in reverse order, we obtain
beginalign*
S_n & = a_n + a_n - 1 + a_n - 2 + cdots + a_3 + a_2 + a_1\
& = [a_1 + (n - 1)d] + [a_1 + (n - 2)d] + [a_1 + (n - 3)d] + cdots + [a_1 + 2d] + [a_1 + d] + a_1
endalign*
Adding the two expressions for $S_n$ yields
beginalignat*10
S_n & = & a_1 & + & [a_1 + d] & + & cdots & + & [a_1 + (n - 1)d] & + & [a_1 + (n - 1)d]\
S_n & = & [a_1 + (n - 1)d] & + & [a_1 + (n - 2)d] & + & cdots & + & [a_1 + d] & + & a_1\ hline
2S_n & = & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d] & + & cdots & + & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d]
endalignat*
Since there are $n$ columns in that sum, we obtain
beginalign*
2S_n & = n[2a_1 + (n - 1)d]\
2S_n & = n[a_1 + a_1 + (n - 1)d]\
2S_n & = n(a_1 + a_n)\
S_n & = fracn(a_1 + a_n)2
endalign*
from which you can find $S_20$ by substituting $4$ for $a_1$ and using the formula $a_n = a_1 + (n - 1)d$ with $n = 20$ and $d = 3$ to find $a_20$.
answered Aug 1 at 9:43
N. F. Taussig
38k93053
add a comment |Â
up vote
1
down vote
You have correctly found the common difference.
The $k$th term of an arithmetic sequence with initial term $a_1$ and common difference $d$ is
$$a_k = a_1 + (k - 1)d$$
Since we are given that $a_3 = 10$ and $a_5 = 16$, we obtain
beginalign*
a_3 & = a_1 + (3 - 1)d = 10\
a_5 & = a_1 + (5 - 1)d = 16
endalign*
which is a system of two linear equations in two variables. Subtracting the first equation from the second yields $2d = 6 implies d = 3$, as you found. Substituting $d = 3$ in the equation for $a_3$ yields $a_1 = 4$, as you found in the comments.
The $n$th partial sum (the sum of the first $n$ terms) of an arithmetic series
beginalign*
S_n & = sum_k = 1^n a_k\
& = sum_k = 1^n [a_1 + (k - 1)d]\
& = sum_k = 1^n a_1 + sum_k = 1^n (k - 1)d\
& = a_1sum_k = 1^n 1 + dsum_k = 1^n (k - 1)\
& = na_1 + fracd(n - 1)n2
endalign*
from which you can find $S_20$ by substituting $20$ for $n$, $3$ for $d$, and $4$ for $a_1$.
Alternatively,
beginalign*
S_n & = sum_k = 1^n a_k\
& = a_1 + a_2 + a_3 + cdots + a_n - 2 + a_n - 1 + a_n\
& = a_1 + [a_1 + d] + [a_1 + 2d] + cdots + [a_1 + (n - 3)d] + [a_1 + (n - 2)d] + [a_1 + (n - 1)d]
endalign*
Since we obtain the same sum if we write the terms in reverse order, we obtain
beginalign*
S_n & = a_n + a_n - 1 + a_n - 2 + cdots + a_3 + a_2 + a_1\
& = [a_1 + (n - 1)d] + [a_1 + (n - 2)d] + [a_1 + (n - 3)d] + cdots + [a_1 + 2d] + [a_1 + d] + a_1
endalign*
Adding the two expressions for $S_n$ yields
beginalignat*10
S_n & = & a_1 & + & [a_1 + d] & + & cdots & + & [a_1 + (n - 1)d] & + & [a_1 + (n - 1)d]\
S_n & = & [a_1 + (n - 1)d] & + & [a_1 + (n - 2)d] & + & cdots & + & [a_1 + d] & + & a_1\ hline
2S_n & = & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d] & + & cdots & + & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d]
endalignat*
Since there are $n$ columns in that sum, we obtain
beginalign*
2S_n & = n[2a_1 + (n - 1)d]\
2S_n & = n[a_1 + a_1 + (n - 1)d]\
2S_n & = n(a_1 + a_n)\
S_n & = fracn(a_1 + a_n)2
endalign*
from which you can find $S_20$ by substituting $4$ for $a_1$ and using the formula $a_n = a_1 + (n - 1)d$ with $n = 20$ and $d = 3$ to find $a_20$.
answered Aug 1 at 9:43
N. F. Taussig
38k93053
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You have correctly found the common difference.
The $k$th term of an arithmetic sequence with initial term $a_1$ and common difference $d$ is
$$a_k = a_1 + (k - 1)d$$
Since we are given that $a_3 = 10$ and $a_5 = 16$, we obtain
beginalign*
a_3 & = a_1 + (3 - 1)d = 10\
a_5 & = a_1 + (5 - 1)d = 16
endalign*
which is a system of two linear equations in two variables. Subtracting the first equation from the second yields $2d = 6 implies d = 3$, as you found. Substituting $d = 3$ in the equation for $a_3$ yields $a_1 = 4$, as you found in the comments.
The $n$th partial sum (the sum of the first $n$ terms) of an arithmetic series
beginalign*
S_n & = sum_k = 1^n a_k\
& = sum_k = 1^n [a_1 + (k - 1)d]\
& = sum_k = 1^n a_1 + sum_k = 1^n (k - 1)d\
& = a_1sum_k = 1^n 1 + dsum_k = 1^n (k - 1)\
& = na_1 + fracd(n - 1)n2
endalign*
from which you can find $S_20$ by substituting $20$ for $n$, $3$ for $d$, and $4$ for $a_1$.
Alternatively,
beginalign*
S_n & = sum_k = 1^n a_k\
& = a_1 + a_2 + a_3 + cdots + a_n - 2 + a_n - 1 + a_n\
& = a_1 + [a_1 + d] + [a_1 + 2d] + cdots + [a_1 + (n - 3)d] + [a_1 + (n - 2)d] + [a_1 + (n - 1)d]
endalign*
Since we obtain the same sum if we write the terms in reverse order, we obtain
beginalign*
S_n & = a_n + a_n - 1 + a_n - 2 + cdots + a_3 + a_2 + a_1\
& = [a_1 + (n - 1)d] + [a_1 + (n - 2)d] + [a_1 + (n - 3)d] + cdots + [a_1 + 2d] + [a_1 + d] + a_1
endalign*
Adding the two expressions for $S_n$ yields
beginalignat*10
S_n & = & a_1 & + & [a_1 + d] & + & cdots & + & [a_1 + (n - 1)d] & + & [a_1 + (n - 1)d]\
S_n & = & [a_1 + (n - 1)d] & + & [a_1 + (n - 2)d] & + & cdots & + & [a_1 + d] & + & a_1\ hline
2S_n & = & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d] & + & cdots & + & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d]
endalignat*
Since there are $n$ columns in that sum, we obtain
beginalign*
2S_n & = n[2a_1 + (n - 1)d]\
2S_n & = n[a_1 + a_1 + (n - 1)d]\
2S_n & = n(a_1 + a_n)\
S_n & = fracn(a_1 + a_n)2
endalign*
from which you can find $S_20$ by substituting $4$ for $a_1$ and using the formula $a_n = a_1 + (n - 1)d$ with $n = 20$ and $d = 3$ to find $a_20$.
answered Aug 1 at 9:43
N. F. Taussig
38k93053
You have correctly found the common difference.
The $k$th term of an arithmetic sequence with initial term $a_1$ and common difference $d$ is
$$a_k = a_1 + (k - 1)d$$
Since we are given that $a_3 = 10$ and $a_5 = 16$, we obtain
beginalign*
a_3 & = a_1 + (3 - 1)d = 10\
a_5 & = a_1 + (5 - 1)d = 16
endalign*
which is a system of two linear equations in two variables. Subtracting the first equation from the second yields $2d = 6 implies d = 3$, as you found. Substituting $d = 3$ in the equation for $a_3$ yields $a_1 = 4$, as you found in the comments.
The $n$th partial sum (the sum of the first $n$ terms) of an arithmetic series
beginalign*
S_n & = sum_k = 1^n a_k\
& = sum_k = 1^n [a_1 + (k - 1)d]\
& = sum_k = 1^n a_1 + sum_k = 1^n (k - 1)d\
& = a_1sum_k = 1^n 1 + dsum_k = 1^n (k - 1)\
& = na_1 + fracd(n - 1)n2
endalign*
from which you can find $S_20$ by substituting $20$ for $n$, $3$ for $d$, and $4$ for $a_1$.
Alternatively,
beginalign*
S_n & = sum_k = 1^n a_k\
& = a_1 + a_2 + a_3 + cdots + a_n - 2 + a_n - 1 + a_n\
& = a_1 + [a_1 + d] + [a_1 + 2d] + cdots + [a_1 + (n - 3)d] + [a_1 + (n - 2)d] + [a_1 + (n - 1)d]
endalign*
Since we obtain the same sum if we write the terms in reverse order, we obtain
beginalign*
S_n & = a_n + a_n - 1 + a_n - 2 + cdots + a_3 + a_2 + a_1\
& = [a_1 + (n - 1)d] + [a_1 + (n - 2)d] + [a_1 + (n - 3)d] + cdots + [a_1 + 2d] + [a_1 + d] + a_1
endalign*
Adding the two expressions for $S_n$ yields
beginalignat*10
S_n & = & a_1 & + & [a_1 + d] & + & cdots & + & [a_1 + (n - 1)d] & + & [a_1 + (n - 1)d]\
S_n & = & [a_1 + (n - 1)d] & + & [a_1 + (n - 2)d] & + & cdots & + & [a_1 + d] & + & a_1\ hline
2S_n & = & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d] & + & cdots & + & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d]
endalignat*
Since there are $n$ columns in that sum, we obtain
beginalign*
2S_n & = n[2a_1 + (n - 1)d]\
2S_n & = n[a_1 + a_1 + (n - 1)d]\
2S_n & = n(a_1 + a_n)\
S_n & = fracn(a_1 + a_n)2
endalign*
from which you can find $S_20$ by substituting $4$ for $a_1$ and using the formula $a_n = a_1 + (n - 1)d$ with $n = 20$ and $d = 3$ to find $a_20$.
answered Aug 1 at 9:43
N. F. Taussig
38k93053
edited Aug 1 at 17:16
answered Aug 1 at 9:43
N. F. Taussig
38k93053
answered Aug 1 at 9:43
N. F. Taussig
38k93053
answered Aug 1 at 9:43
N. F. Taussig
38k93053
38k93053
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1
You've done well so far. If the common difference is $3$ and the third term is $10$, what is the second term? Once you answer that, you'll find the rest is easy.
– saulspatz
Jul 31 at 15:11
2
@saulspatz Oh! So, the second term is 7 and the first term would be 4.
– Ella
Jul 31 at 15:15
There you go, that's it!
– saulspatz
Jul 31 at 15:18